Question: You have found the following ages (in years) of 4 zebras. Those zebras were randomly selected from the 45 zebras at your local zoo: $ 14,\enspace 15,\enspace 9,\enspace 1$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 45 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\overline{x}} = \dfrac{14 + 15 + 9 + 1}{{4}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {17.64} + {27.04} + {0.64} + {77.44}} {{4 - 1}} $ {s^2} = \dfrac{{122.76}}{{3}} = {40.92\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{40.92\text{ years}^2}} = {6.4\text{ years}} $ We can estimate that the average zebra at the zoo is 9.8 years old. There is also a standard deviation of 6.4 years.